A curve in the plane is defined parametrically by the equations $x=\ln(4t-3)$ and $y=\dfrac{2}{t}$. Find the value of $\dfrac{dy}{dx}$ at $t=3$. Choose 1 answer: Choose 1 answer: (Choice A) A $-\dfrac12$ (Choice B) B $-6$ (Choice C) C $-\dfrac14$ (Choice D) D $\dfrac{2}{3}$
In general, to find the derivative (i.e. the expression for $\dfrac{dy}{dx}$ ) of a function defined parametrically by the equations $x=u(t)$ and $y=v(t)$ (where $u$ and $v$ are any functions of $t$ ), we use the following rule: $\dfrac{dy}{dx}=\dfrac{\left(\dfrac{dy}{dt}\right)}{\left(\dfrac{dx}{dt}\right)}=\dfrac{v'(t)}{u'(t)}$ We are given that $x=\ln(4t-3)$ and $y=\dfrac{2}{t}$ : $\begin{aligned} \dfrac{dy}{dx}&=\dfrac{\dfrac{d}{dt}\left(\dfrac{2}{t}\right)}{\dfrac{d}{dt}(\ln(4t-3))} \\\\ &=\dfrac{\left(\dfrac{-2}{t^2}\right)}{\left(\dfrac{4}{4t-3}\right)} \\\\ &=\dfrac{3-4t}{2t^2} \gray{\text{Simplify}} \end{aligned}$ Now let's evaluate $\dfrac{dy}{dx}$ at $t= 3$ : $\begin{aligned} &\phantom{=}\dfrac{3-4( 3)}{2( 3)^2} \\\\ &=\dfrac{-9}{18} \\\\ &=-\dfrac12 \end{aligned}$ In conclusion, the value of $\dfrac{dy}{dx}$ at $t=3$ is $-\dfrac12$.